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3 years ago

Two hockey pucks, each with a mass of 0.2 kg, slide across the ice and

Physics

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

Answe: It's B (apex)

Explanation:

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A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

svlad2 [7]

Force=mass*acceleration
F=ma
F=25*5
F=100 N

8 0

3 years ago

V=xf-xi/t solve for t

irina [24]

Answer:

$t=\frac{x_f-x_i}{v}$

Explanation:

Starting from the equation:

$v=\frac{x_f-x_i}{t}$

First of all, let's multiply by t on both sides:

$v\cdot t = \frac{x_f-x_i}{t}\cdot t \\vt = x_f - x_i$

And then, let's divide by v on both sides:

$\frac{vt}{v}=\frac{x_f-x_i}{v}\\t=\frac{x_f-x_i}{v}$

So, finally

$t=\frac{x_f-x_i}{v}$

7 0

3 years ago

At what point does the external energy enter the system?

Phoenix [80]

The correct answer as the first one above !

8 0

3 years ago

A vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement

Leto [7]

Answer:

|x| = √53

Explanation:

We are told that the vector starts at the point (0.0) and ends at (2,-7) .

Thus, magnitude of displacement is;

|x| = √(((-7) - 0)² + (2 - 0)²)

|x| = √(49 + 4)

|x| = √53

5 0

3 years ago

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

andrew-mc [135]

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm$

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0

3 years ago